Find derivative:

Question

y= e^(x^x)

Answer 1

if y = u^v then
y' = vu^(v-1)u' + lnu u^v v'
Note that u^n and a^u are just special cases of this
So, what you have is
y= e^u
y' = e^u u'
u = x^x, so u' = x^x (1+lnx)

y' = e^(x^x) x^x (1+lnx)

Answer 2

let z = x^x

y = e^z
dy/dz = z e^z = x^x (e^x^x)

dy/dx = dy/dz * dz/dx so we need dz/dx when z = x^x

look here
http://www.analyzemath.com/calculus/Differentiation/first_derivative.html for
dz/dx = (ln x + 1) x^x
so
x^x ( e^x^x) (ln x + 1) x^x
=x^2x e^x^x (ln x + 1)

Answer 3

where did you get

d/dz e^z = z e^z?